Weijia Li
Econ 460
Feb 21st, 2022
Problem Set 4
1. US import 1947-2021
a. Plot the imports and its natural log of the United States 1947-2000
b. The import series is better represented using the exponential trend.
c. The exponential trend of the estimated import from 1947 to 2000.
d. Generate the point and 90 percent interval estimate for the log level of imports
from 2001 to 2020.
e. Plotting the forecast against the actual, I find that the forecast is overall reliable
because the actual values fall into the 90 percent interval most of time. The actual
imports dropped drastically in the year of 2009 and 2020, most likely due to the
financial crisis and COVID-19 pandemic. In other periods of time, the estimation
is accurate.
f. Generate point and 90 percent interval estimate of level of imports from 2001
to 2021.
g. This creates an exponential trend of estimate, and the graph shows that the
estimate is highly accurate, with two deviations in the year of 2009 and 2020.
Similar to the graph in question e, these outliers are largely the result of the
financial crisis and the pandemic.
h. Re-estimate using the full sample 1947 − 2021Q4. Generate point and 90%
interval forecasts for the level of imports for the next 20 quarters.
i. This forecast does not appear to be reliable because the actual level of imports
has increased in a slower speed whereas the exponential estimate predicts that
the import would increase faster. After the shocks in imports caused by the
financial crisis and the pandemic, the actual increase in level of imports fall
behind the estimate substantially. Therefore, the estimate appears to be not
reliable.
2. Time series
𝑦𝑡 = 𝛽0 + 𝛽1𝑇𝑖𝑚𝑒𝑡 + 𝜀𝑡
𝐸(𝑌𝑡 ) = 𝛽̂ + 𝛽̂1 (𝑡𝑖𝑚𝑒𝑡 )
𝛽̂0 = 0.51𝛽̂1 = 0.02𝜎̂ 2 = 16
T=100
T=101
T=102
T=103
T=104
a. Point Estimate:
2021q4
2022q1
2022q2
2022q3
2022q4
90 Percent Interval Estimate:
2021q4
2022q1
2022q2
2022q3
2022q4
b. 𝑦𝑡 = 𝑙𝑛(𝑌𝑡 )
2021q4
2022q1
2022q2
2022q3
2022q4
𝑦100
𝑦101
𝑦102
𝑦103
𝑦104
= 0.51 + 0.02 × 100 = 2.51
= 0.51 + 0.02 × 101 = 2.53
= 0.51 + 0.02 × 102 = 2.55
= 0.51 + 0.02 × 103 = 2.57
= 0.51 + 0.02 × 104 = 2.59
[2.51 − 1.645 × 4, 2.51 + 1.645 × 4] = [−4.07, 9.09]
[2.53 − 1.645 × 4, 2.51 + 1.645 × 4] = [−4.05, 9.11]
[2.55 − 1.645 × 4, 2.51 + 1.645 × 4] = [−4.03, 9.13]
[2.57 − 1.645 × 4, 2.51 + 1.645 × 4] = [−4.01, 9.15]
[2.59 − 1.645 × 4, 2.51 + 1.645 × 4] = [−3.99, 9.17]
𝑌𝑡 = 𝑒 𝑦𝑡
𝑌104 = 𝑒 2.59 = 13.32977
90 Interval Forecast: [𝑒 −3.99 , 𝑒 9.17 ] = [0.01850, 9604.6247]
3. 𝑇𝑡 = 𝛽0 + 𝛽1 𝑇𝑖𝑚𝑒𝑡 , 𝑠𝑢𝑝𝑝𝑜𝑠𝑒 𝑡ℎ𝑎𝑡 𝛽1 > 0
a. We are sure that this series is expected to grow in subsequent periods because
𝛽1 > 0. This is a linear equation, so it would grow in the future periods.
b. We are unsure whether this series will grow with certainty in every period. We do
not know the variance of the estimation. In most cases, the forecast estimate
become less reliable because of changes in actual growth that are not presented in
the time series.